Problem: Is ${463377}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {463377}= &&{4}\cdot100000+ \\&&{6}\cdot10000+ \\&&{3}\cdot1000+ \\&&{3}\cdot100+ \\&&{7}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {463377}= &&{4}(99999+1)+ \\&&{6}(9999+1)+ \\&&{3}(999+1)+ \\&&{3}(99+1)+ \\&&{7}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {463377}= &&\gray{4\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {4}+{6}+{3}+{3}+{7}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${463377}$ is divisible by $3$ if ${ 4}+{6}+{3}+{3}+{7}+{7}$ is divisible by $3$ Add the digits of ${463377}$ $ {4}+{6}+{3}+{3}+{7}+{7} = {30} $ If ${30}$ is divisible by $3$ , then ${463377}$ must also be divisible by $3$ ${30}$ is divisible by $3$, therefore ${463377}$ must also be divisible by $3$.